Solve: $tan^{-1} \left( \dfrac{x-1}{x-2}\right)$ + $tan^{-1} \left( \dfrac{x+1}{x+2}\right)$ = $\dfrac{\pi}{4}$

The correct answer is $x= \pm {\dfrac {1}{\sqrt2}}$

$tan^{-1} \left( \dfrac{x-1}{x-2}\right)$ + $tan^{-1} \left( \dfrac{x+1}{x+2}\right)$ = $\dfrac{\pi}{4}$

Using the formula $tan^{-1} \left( A\right)$ + $tan^{-1} \left( B \right)$ = $tan^{-1} \left( \dfrac{A+B}{1-AB}\right)$, we can write the above equation as follows

$tan^{-1} \left( \dfrac{\dfrac{x-1}{x-2} + \dfrac{x+1}{x+2}}{1- \dfrac{x-1}{x-2} \dfrac{x+1}{x+2}} \right)$ = $\dfrac{\pi}{4}$


⇒ $\dfrac{(x-1)(x+2) + (x+1)(x+2)}{x^2 - 4 -x^2 + 1}$ = $tan\dfrac{\pi}{4} = 1$

⇒ $\dfrac{x^2+2x-x-2 + x^2-2x+x-2}{x^2 - 4 -x^2 + 1}$ = $1$

⇒ $2x^2-4=-3$

⇒ $x= \pm \sqrt {\dfrac {1}{2}}$

⇒ $x= \pm {\dfrac {1}{\sqrt2}}$